Energy and Power Problems and Solutions

Problem#1

The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms. The total mass of the train is 875 g. Find the average power delivered to the train during the acceleration.

Answer:

Given: v = 0.625 m/s, ∆t = 21.0 ms = 21.0 x 10-3 s and m = 875 g = 0.875 kg, then with

Pav = W/∆t = ∆K/∆t

We get the average power delivered to the train during the acceleration is

Pav = Kf/∆t = ½ mv2/∆t

Pav = ½ (0.875 kg)(0.620 m/s)2/(21 x 10-3 s) = 8.01 W

Problem#2

A 700-N Marine in basic training climbs a 10.0-m vertical rope at a constant speed in 8.00 s. What is his power output?

Answer:

Given: h = 10.0 m, mg = 700 N, and ∆t = 8.00 s, then with

Pav = W/∆t = mgh/∆t

We get

Pav = (700 N)(10.0 m)/8.00 s = 875 W

Problem#3

Make an order-of-magnitude estimate of the power a car engine contributes to speeding the car up to highway speed. For concreteness, consider your own car if you use one. In your solution state the physical quantities you take as data and the values you measure or estimate for them. The mass of the vehicle is given in the owner’s manual. If you do not wish to estimate for a car, consider a bus or

truck that you specify.

Answer:

A 1 300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s.

The output work of the engine is equal to its final kinetic energy,

Kf = ½ mvf2 = ½ (1300 kg)(24.6 m/s)2 = 390 kJ

With power

P = Kf/∆t = 390 kJ/15.0s = 26 kW

P = 26 kW x 0.00134 hp/W = 34.8 horsepower

Problem#4

A skier of mass 70.0 kg is pulled up a slope by a motordriven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30.0° slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform this task?

Answer;

(a) ∑W = ∆K , but ∆K = 0 because he moves at constant speed.

The skier rises a vertical distance of 60 m sin30.00 = 30.0 m. Thus,

Win = –Wg = (70.0 kg)(9.80 m/s2)(30.0 m) = 20.6 kJ

(b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,

Pin = W/∆t = 20.6 kJ/30.0s

Pin = 686 W = 0.919 hp

Problem#5

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this power compare with the motor power when the elevator moves at its cruising speed?

Answer:

(a) The distance moved upward in the first 3.00 s is

∆y = vavgt = ½ (0 + 1.75 m/s)(3.00s) = 2.63 m

The motor and the earth’s gravity do work on the elevator car:

∑W = ∆K

Wmotor + mg∆y cos1800 = Kf – Ki

Wmotor + (650 kg)(9.80 m/s2)(2.63 m)(–1) = ½ (650 kg)(1.75 m/s)2 – 0

Wmotor = 995 J + 1.68 x 103 J = 1.77 x 104 J

So,

Pavg = W/∆t = 1.77 x 104 J/3.00s = 5.91 kJ = 7.92 hp

(b) When moving upward at constant speed (v = 1.75 m/s), therefore

P = F.v = (650 kg)(9.80 m/s2)(1.75 m/s) = 11.1 kW = 14.9 hp 

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